What is the integral from 0 to 24/25 of sqrt(1+25x) ?

The integral from 0 to 24/25 of sqrt(1+25x) is 248/75

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integral from 0 to 24/25 of sqrt(1+25x)

\int_{0}^{\frac{24}{25}} 1 + 25 x d x

\frac{248}{75}

Decimal

3.30666 \dots

Solution steps

\int_{0}^{\frac{24}{25}} 1 + 25 x d x

Apply u-substitution

: \int_{1}^{25} \frac{u}{25} d u

= \int_{1}^{25} \frac{u}{25} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{25} \cdot \int_{1}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

= \frac{1}{25} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25}

Compute the boundaries:

\frac{248}{3}

= \frac{1}{25} \cdot \frac{248}{3}

Simplify

= \frac{248}{75}

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What is the integral from 0 to 24/25 of sqrt(1+25x) ?
The integral from 0 to 24/25 of sqrt(1+25x) is 248/75